Given a geodesic equation in a quantum mechanical context. How does one make a relation to the field equation?

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Question
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Given a geodesic equation in a quantum mechanical context. How does make a relation to the field equation? (Or how does one guess the relation between curvature and mass)?

Motivation
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Proceeding from an equivalence principle directly to the geodesic equation is known. In the same spirit we use the Heisenberg picture to define velocity $\hat v$:

$$\hat v = \frac{dU^\dagger x U}{dt} = U^\dagger\frac{[H,x]}{- i \hbar}U$$

where $U^\dagger$ is the unitary operator and $H$ is the Hamiltonian.

Now we can again differentiate to get acceleration $\hat a$:

$$\hat a = \hat U^\dagger\frac{[[\hat H, \hat x], \hat x]}{-i \hbar} \hat U = \hat U^\dagger\frac{(\hat H^2 \hat x + \hat x \hat H^2 - 2\hat H \hat x \hat H)}{\hbar^2} \hat U$$

We can simplify the calculation by splitting the Hamiltonian into potential  $\hat V$ and kinetic energy $\hat T$: $\hat H = \hat T + \hat V$

By noticing (one can also calculate this) that the acceleration of an object in a constant potential is $0$:

$$\hat 0 = \hat T^2 x + x \hat T^2 - 2 \hat T \hat x \hat T$$

We also know $[\hat V, \hat x] = 0$ as potential is a function of position. Thus, we can simplify acceleration as:

$$\hat a \hbar^2 = \hat V \hat T \hat x + \hat x \hat T \hat V - \hat V \hat x \hat T - \hat T \hat x \hat V$$

Note this acceleration operator also commutes with position. Also acceleration only makes sense in Quantum Mechanics and therefore we are in the regime $v\ll c$:

$$[\hat a , \hat x ]=0$$
To make this physical law in uniformity with general covariance. We replace the $\nabla^2$ with the Laplace–Beltrami operator:

$$\hat T = \frac{1}{2m} \Delta_{LB}\ = \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big)$$

And redefine $r$ to be a three vector in any co-ordinate system (for example in Cartesian coordinates $\hat r = (x,y,z)$ and $\hat a = (a_x , a_y, a_z$):

$$\hat a \hbar^2 = \hat V \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big) \hat r + \hat r \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big)\hat V$$
$$- \hat V \hat r \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big) - \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big) \hat r \hat V$$

We will now proceed with the final step of this derivation. Suppose that no particles are accelerating in the neighbourhood of a point-event with respect to a freely falling coordinate system:

$$0 = \hat V \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big) \hat r + \hat r \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big)\hat V$$
$$- \hat V \hat r \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big) - \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big) \hat r \hat V$$

Hence, given a metric in the limit $c \to \infty$ will follow the geodesic equation above. One would have to solve for the potential $V$ and substitute that in $H\psi = (T + V)\psi$

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First of all, I recommend to inform yourself about very well known results before engaging in speculative arguments. Topics to look up would be relativistic quantum mechanics and its challenges (typically included as an introductory chapter in most quantum field theory books), and a course on general relativity where you can see how the Hamiltonian of a geodesic looks and how the Newtonian limit of geodesic motion is obtained. From that point it should be easy to go to what I describe below.

The governing equation for the quantum states of a spin-0 particle is

$$(g^{\mu\nu} \partial_\mu \phi \sqrt{-g})_{,\nu} + m^2 \phi =0$$

This equation corresponds to a particle which moves on geodesics in the classical approximation. (You can derive this through the Eikonal approximation.)

When you go to the Newtonian limit of this equation (which can be a somewhat nontrivial procedure depending on how rigorous you want to be), you will obtain the Schrödinger equation coupled to a Newtonian gravitational field

$$i \hbar \partial_t \psi = (-\frac{\hbar^2}{2m} \Delta + m \Phi) \psi$$

where $\Phi$ is the usual Newtonian gravitational potential and we get rid of a "Zitterbewegung" term by a field redefinition  $\psi = e^{im c^2t} \phi$.

answered Aug 15, 2017 by (1,620 points)

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